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Good day people

I am modelling a water bottle rocket. Using the conservation of mass :

$$-{\rho}vA + \frac{d}{dt}∫dM = 0 \tag{1}$$

Since the mass, O2 pressure, O2 volume and velocity change over time :

$$ \rho = \frac{M}{V} \tag{2} $$

and $$PV = MRT $$ R in J/kg.K : Assume constant T $$ M = PV/RT \tag{3} $$

Plugging (2) and (3) into (1) we find :

$$ \frac{d}{dt} ∫d(\frac{PV}{RT}) = \frac{PVvA}{RTV} $$

Since $RT$ is constant :

$$ \frac{1}{RT}\frac{d}{dt}∫d(PV) = \frac{PvA}{RT} $$

Simplifying ( sort of )

$$ \frac{1}{P}\frac{d}{dt}∫d(PV) = vA$$

This is where is start to doubt my math :

$$\frac{d}{dt}∫\frac{1}{P}d(PV) = vA $$ : Can I put P inside the integral?

Assuming I can. $$d(PV) = PdV + VdP $$using chain rule

thus

$$ \frac{d}{dt}∫(\frac{PdV}{P} + \frac{VdP}{P}) = vA$$

after integrating :

$$\frac{d}{dt}(V_1-V_0 + Vln(\frac{P_1}{P_0}) = vA $$: Where I have values for V0 and P0. but I do not know what the value for V is.( coefficient of the ln term)

multiplying by dt on both sides :

$$ d(V_1-V_0 + Vln(\frac{P_1}{P_0}) = vA dt $$: And this is where I am stuck. Substituting my values : $V_1 = 0.057 ; V_0 = 0.848 ; P_1 = 101.3 ; P_0 = 1500 $

$$ d(0.791 - 2.69V) = vA dt $$
integrating : $$ ∫d(0.791 - 2.69V) = ∫vA dt $$

$$ 0.791 -2.69(V-V_0) = A∫vdt $$

simplifying

$$ V = \frac{A}{-2.69}∫vdt - \frac{0.791}{-2.69}+V_0 $$

$$ V = -0.105∫vdt-0.294+0.057 $$

$$ V = -0.105∫vdt-0.237 $$

Does this seem correct?

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  • $\begingroup$ For the integral $\int dM$, what are you integrating over? In any case, you certainly cannot take $P$ inside in the integral, unless $P$ is constant on the domain over which you integrate. And your change of variable using $d(PV) = PdV + VdP$ is suspect: you are absorbing a lot of details in your notation (the domain of integration, for example, needs to be changed when you change variables). Hmmm... it will help a lot if way up top at the beginning you define all your symbols. I can see that $R$ is just a constant, and $T$ is the temperature which apparently we can set to a constant... $\endgroup$ – Willie Wong Feb 20 '14 at 11:55
  • $\begingroup$ ... $P$ is presumably some sort of pressure (but of what?), and $M$ is the mass of something (what is it?). $\rho$ looks like a density of whatever $M$ is, and $V$ a certain volume (what's the volume measuring?). But $A$ and $v$ I have no idea... $\endgroup$ – Willie Wong Feb 20 '14 at 11:56
  • $\begingroup$ I found that I have done this wrong. As the rocket propels, it loses mass, and the air pressure decreases. But, the mass of air is constant, and I designed the rocket that once all the water is expelled, the air left inside the rocket will be at atmospheric pressure. What I did here^ was model the volume change of the air inside the tank as it leaves the rocket. But it doesn't, the air inside the tank, stays inside the tank. I should just have done the mass balance on the water. I did it before, but doubted my math after I did an example in our handbook. So please, Dis regard this post. $\endgroup$ – 22134484 Feb 20 '14 at 15:05
  • $\begingroup$ I will post a new topic. I will add everything I have done up to this point to give background on what I have and what I need eventually. The post will be named : Modelling a Water Rocket. Hope to see some of you there, especially Willie Wong. $\endgroup$ – 22134484 Feb 20 '14 at 15:07
  • $\begingroup$ follow up topic is math.stackexchange.com/questions/684191 $\endgroup$ – 22134484 Feb 21 '14 at 0:26

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